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Math Help - Sketch the graph

  1. #1
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    Sketch the graph

    Please help me sketch this graph:


    f(x)=0 for x=0
    f'(x)>0 for x<-2
    f'(x)=0 for x>2
    f'(x)=0 for x=-2
    f'(x)<0 for -2<x<2
    f"(x)<0 for x<1
    f"(x)=0 for x=1
    f" (x)>0 for 1<x<2

    Thanks
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  2. #2
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    Quote Originally Posted by qbkr21 View Post
    Please help me sketch this graph:


    f(x)=0 for x=0
    f'(x)>0 for x<-2
    f'(x)=0 for x>2
    f'(x)=0 for x=-2
    f'(x)<0 for -2<x<2
    f"(x)<0 for x<1
    f"(x)=0 for x=1
    f" (x)>0 for 1<x<2

    Thanks
    I don't know how to draw in computers so I cannot show the sketch. But I can describe it.

    f(x)=0,for x=0
    That means the graph passes point (0,0).

    f'(x)=0 for x=-2
    That means the graph is horizontal at x = -2.

    f"(x)=0 for x=1
    That means the concavity of the graph changes at x = 1. There is an inflection point at x=1. The graph there needs not be horizontal, though.

    f'(x)>0 for x<-2
    The slope of the graph is positive to the left of x = -2.
    So if at x = -2, the graph is horizontal, then there is maximum point at x = -2.

    f'(x)<0 for -2<x<2
    From x = -2 up to x = 2, the slope is negative.

    f'(x)=0 for x>2
    The graph is horizontal to the right of x = 2. Horizontal up to before infinity.

    f"(x)<0 for x<1
    The concavity of the graph is downwards to the left of x = 1.
    That means from negative infinity up to x=1, the graph is like a vertical parabola that opens downward, whose vertex is at x = -2.

    f" (x)>0 for 1<x<2
    The concavity of the graph is upwards between x=1 and x=2. Yet, the slope there is still negative.

    ---------------------
    Okay, if you are still with me, the graph is like a weak or tipsy number two.
    From after negative infinity up to x=1 it looks like a parabola with its highest point at x = -2.
    It passes through the origin (0,0), so the highest point of the "parabola" is above the x-axis.
    Then at x=1 it changes concavity--it faces upward, but it is still going down.
    Then at x=2 it becomes horizontal until it reaches very close to positive infinity.
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  3. #3
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    Hello, qbkr21!

    I totally agree with ticbol's analysis . . .


    Sketch the graph:

    f(x) = 0

    \begin{array}{cccc}f'(x) > 0 & \text{for }x<-2 \\<br />
f'(x) < 0 & \text{-}2 < x < 2 \\<br />
f'(x) = 0 & \text{for }x = 2 \\<br />
f'(x)=0 & \text{for }x>2\end{array}

    \begin{array}{ccc}f"(x)<0 &\text{for }x<1 \\<br />
f"(x)=0 & \text{for }x=1 \\<br />
f"(x)>0 & 1<x<2 \end{array}

    Take the statements one at a time . . .

    f(0) = 0: the graph contains the origin.


    f'(x) > 0\text{ for }x < -2
    . . To the left of x = -2, the graph is rising: \nearrow

    f'(x) < 0\text{ for }\text{-}2 < x < 2
    . . Between -2 and +2, the graph is falling: \searrow

    f'(2) = 0
    . . At x = 2, there is a horizontal tangent: \rightarrow

    f'(x) = 0\text{ for }x > 2
    . . To the right of x = 2, the graph is a horizontal line: \rightarrow


    f''(x) < 0\text{ for }x < 1
    . . To the left of x = 1, the graph is concave down: \cap

    f''(1) = 0
    . . At x = 1, there is an inflection point.

    f''(x) > 0\text{ for }1 < x < 2
    . . From x = 1 to x = 2, the graph is concave up: \cup


    The graph looks something like this:
    Code:
                          |
                          |
            *             |
        *   :     *       |
      *     :         *   |     1     2
      - - - + - - - - - - * - - + - - + - - - - -
           -2             |  *  :     :
                          |    *:     :
                          |     *     :
                          |       *   :
                          |           *  *  *  *
                          |

    Evidently, there is a horizontal tangent at x = -2,
    . . but they didn't mention it.

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  4. #4
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    Quote Originally Posted by Soroban View Post
    Hello, qbkr21!

    I totally agree with ticbol's analysis . . .



    Take the statements one at a time . . .

    f(0) = 0: the graph contains the origin.


    f'(x) > 0\text{ for }x < -2
    . . To the left of x = -2, the graph is rising: \nearrow

    f'(x) < 0\text{ for }\text{-}2 < x < 2
    . . Between -2 and +2, the graph is falling: \searrow

    f'(2) = 0
    . . At x = 2, there is a horizontal tangent: \rightarrow

    f'(x) = 0\text{ for }x > 2
    . . To the right of x = 2, the graph is a horizontal line: \rightarrow


    f''(x) < 0\text{ for }x < 1
    . . To the left of x = 1, the graph is concave down: \cap

    f''(1) = 0
    . . At x = 1, there is an inflection point.

    f''(x) > 0\text{ for }1 < x < 2
    . . From x = 1 to x = 2, the graph is concave up: \cup


    The graph looks something like this:
    Code:
                          |
                          |
            *             |
        *   :     *       |
      *     :         *   |     1     2
      - - - + - - - - - - * - - + - - + - - - - -
           -2             |  *  :     :
                          |    *:     :
                          |     *     :
                          |       *   :
                          |           *  *  *  *
                          |

    Evidently, there is a horizontal tangent at x = -2,
    . . but they didn't mention it.


    Good morning, Soroban.

    The 4th statement is actually, "f'(x) = 0 for x = -2". Not for x = 2. So the horizontal tangent at x = -2 is mentioned. That's why I said the vertex of the "parabola" is at x = -2.

    The pesky negative sign bit you.
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  5. #5
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    Hello, ticbol!

    The pesky negative sign bit you. :)
    Ouch!

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